Find the maximum and minimum of F (x) = (SiNx + 1) (cosx + 1)

Find the maximum and minimum of F (x) = (SiNx + 1) (cosx + 1)

f(x)=sinx cosx+sinx+cosx+1
Let t = SiNx + cosx
Then T ^ 2 = (SiNx + cosx) ^ 2 = 1 + 2sinx cosx
^2 is the square
sinx cosx = (t^2-1)/2
f(x)=(t^2-1)/2+t+1
=(t^2)/2+t+1/2
t=sinx+cosx=√2[(√2/2)sinx+(√2/2)cosx]
=√ 2 [sin (x + 45 degrees)]
So the range of T is: [- √ 2, √ 2]
The symmetry axis of F (x) = (T ^ 2) / 2 + T + 1 / 2 is - 1
So when t = - 1, we get the minimum value, 1 / 2-1 + 1 / 2 = 0
When t = √ 2, the maximum value is 2 / 2 + √ 2 + 1 / 2 = 3 / 2 + √ 2